Re: SSTC idea

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Terry Fritz" <teslalist-at-qwest-dot-net>

Hi,

The real problem with a solid state spark gaps is simple economics and 
simplicity.

Why spend thousand of dollars on a complex device few of use could build 
when a simple old motor and disk full of bolts can do the job.

A fully solid state 2000 amp 25kV  spark gap is not technically very 
impossible.  It just cost too much to design build and work out the bugs 
(painful).  I think it could be done for about 2 to 5 thousand dollars and 
one heck of a lot of work.  But right now it is just too exotic to be 
useful to everyone.

Solid state devices like high current more than high voltage.  MOT systems 
will be  the first to go solid state simply because the voltages are 
lower.  25kV will need some major new solid state devices introduced before 
they become practical.  But it is just a mater of time...

Cheers,

         Terry


At 07:33 PM 3/15/2003 -0800, you wrote:
> >
> > >However, the more exciting direction to go is to replace the traditional
> > >spark gap with a solid state version.  This would be a definite challenge
> >
> > I always wonder about this. Would a solid-state gap replacement be any
>more
> > efficient than an old-fashioned spark gap? I guess they use them in radar
> > transmitters for repeatability/jitter reasons but what's the point in a
>TC?
>
>
>Hard to beat a spark gap for high peak power switching...
>
>The peak power you need to switch in a disruptive coil is very, very high
>(far more than any reasonable tube size)..
>
>Take some example numbers...  100 kHz resonant frequency, 20 kV voltage, 120
>bps, 1200 Watts.
>First, what's the joules/bang.. 1200Watts/120 bangs/sec = 10 Joule/bang, so
>now we know that the capacitor is big enough to hold 10 Joules at 20 kV..
>Using E = 1/2 * C * V^2 >>> 20 = C * 400, for V in kV and C in uF.. C =
>20/400 = 0.05 uF.. (50 nF)
>Now, figure out L to get the fres.. 628E3 = 1E6*sqrt(1/(L*0.05))  (for L&C
>in uH and uF) 1/(.628^2)/.05 = L >> L=51 uH
>Now figure out peak primary current (assuming energy in L = energy in C)..
>10 = 1/2*L*I^2 >> 20/sqrt(51E-6) = I... 2.8 kA
>
>so, you need a switch that can switch 3 kA -at- 20 kV.. pretty darn impressive
>switch (it's not quite 60 MW peak power, for a variety of reasons, but
>still.. pretty good)
>
>(actual power >> 10 Joules/ 2.5 microseconds (quarter cycle)  around 4 MW)
>
>Not a heck of a lot of tube or solid state switches that can handle that
>kind of peak power. (yep, those mondo thyratrons they use at SLAC might be
>up in that range..).. and especially not for the  $10-20 that a rudimentary
>sparkgap (that CAN switch that kind of power) costs.
>
>Herein lies the fundamental difference between CW (tube and SSTC) coils and
>disruptive coils.  A huge tube coil (even a 100 kW one) doesn't have the
>peak powers that a disruptive coil does. And that's why the spark
>characteristics are so different.  The power levels when the spark is
>growing are radically different.
>
>
>By lowering the fres to something lower, you can get the peak currents down
>(L gets big) for the same bang energy, but that effectively increases the
>time over which the energy is transferred, reducing peak power to something
>reasonable.. Go far enough, and the peak to average ratio starts to get
>close to 1 and you have a CW coil.
>
>Staccato operation of a tube coil is sort of in between.. you're going for a
>high peak/average ratio.. given that the tube has a long thermal time
>constant, and can probably support a much higher peak power than average.
>The same idea lets you build 1 MW peak power radar magnetrons that aren't
>much bigger than the 600W CW unit in your microwave oven.
> >
> > >I believe though, that a lot of the OLTC (Off-line Tesla Coils) use a
> > >similar solid state switching arrangement however it is at low voltage
> > >(<1kV).
> >
> > That's how mine works at any rate. It's just a DC resonant coil with a
>pair
> > of IGBTs dropped in pace of the spark gap.
> >
> > Steve C.
> >